Question

Consider an infinite geometric series with the first term $a$ and the common ratio $r$. If its sum is $4$ and the second term is $\frac{3}{4}$ then $(a,r)$ has the value

• A. $(\frac{7}{4},\frac{3}{7})$
• B. $(1,\frac{3}{4})$
• C. $(\frac{3}{2},\frac{1}{2})$
• D. $(3,\frac{1}{4})$

Answer 1

Answer: (B), (D)

The correct options are
B $(1,\frac{3}{4})$
D $(3,\frac{1}{4})$

Infinite sum of G.P is given by $\frac{a}{1-r}=4$
and second term is given to be $\frac{3}{4}$
hence $ar=\frac{3}{4}$
so let us solve the above two equations :
$a=\frac{3}{4r}\Rightarrow \frac{3}{4r(1-r)}=4\Rightarrow 16r(1-r)=3\Rightarrow 16r-16r^2=3\Rightarrow 16r^2-16r+3=0\Rightarrow (4r-1)(4r-3)=0\Rightarrow r=\frac{1}{4},\frac{3}{4}$
and $\Rightarrow a=\frac{3}{4r}=3,1$
so the ordered pair $(a,r)=(3,\frac{1}{4}),(1,\frac{3}{4})$
None of the option match